An examination consists of two papers, Paper-1 and Paper-2. The probability of failing in Paper-1 is 0.3 and that in Paper-2 is 0.2. Given that a student has failed in Paper-2, the probability of failing in Paper-1 is 0.6. Find the probability of a student failing in both the papers.

An examination consists of two papers, Paper-1 and Paper-2. The probability of failing in Paper-1 is 0.3 and that in Paper-2 is 0.2. Given that a student has failed in Paper-2, the probability of failing in Paper-1 is 0.6. The probability of a student failing in both the papers is 

(A) 0.5 

(B) 0.18 

(C) 0.12 

(D) 0.06 

Answer: (C) 0.12 

Explanation: 

Let $E_1$ and $E_2$ be the two events of failing in Paper-1 and Paper-2 respectively. 

Given, $P(E_1) = 0.3$, $P(E_2) = 0.2$ and $P(E_1 | E_2) = 0.6$ 

Now, $P(E_1 | E_2) = \dfrac{P(E_1 \cap E_2)}{P(E_2)}$ 

$\Rightarrow 0.6 = \dfrac{P(E_1 \cap E_2)}{0.2}$ 

$\Rightarrow P(E_1 \cap E_2) = 0.12$, which is the required probability.